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Mysterious number 6174

The number 6174 is a really mysterious number. At first glance, it might not seem so obvious. But as we are about to see, anyone who can subtract can uncover the mystery that makes 6174 so special.

Kaprekar's operation

In 1949 the mathematician D. R. Kaprekar from Devlali, India, devised a process now known as Kaprekar's operation. First choose a four digit number where the digits are not all the same (that is not 1111, 2222,...). Then rearrange the digits to get the largest and smallest numbers these digits can make. Finally, subtract the
smallest number from the largest to get a new number, and carry on repeating the operation for each new number.

It is a simple operation, but Kaprekar discovered it led to a surprising result. Let's try it out, starting with the number 2005, the digits of last year. The maximum number we can make with these digits is 5200, and the minimum is 0025 or 25 (if one or more of the digits is zero, embed these in the left hand side of the minimum number). The subtractions are:

When we reach 6174 the operation repeats itself, returning 6174 every time. We call the number 6174 a kernel of this operation. So 6174 is a kernel for Kaprekar's operation, but is this as special as 6174 gets? Well not only is 6174 the only kernel for the operation, it also has one more surprise up its sleeve. Let's try again starting with a different number, say 1789.

9871 - 1789 = 8082
8820 - 0288 = 8532
8532 - 2358 = 6174

We reached 6174 again!

A very mysterious number...

When we started with 2005 the process reached 6174 in seven steps, and for 1789 in three steps. In fact, you reach 6174 for all four digit numbers that don't have all the digits the same. It's marvellous, isn't it? Kaprekar's operation is so simple but uncovers such an interesting result. And this will become even more intriguing when we think about the reason why all four digit numbers reach
this mysterious number 6174.

Only 6174?

The digits of any four digit number can be arranged into a maximum number by putting the digits in descending order, and a minimum number by putting them in ascending order. So for four digits a,b,c,d where

9 ≥ a ≥ b ≥ c ≥ d ≥ 0

and a, b, c, d are not all the same digit, the maximum number is abcd and the minimum is dcba.

We can calculate the result of Kaprekar's operation using the standard method of subtraction applied to each column of this problem:

a

b

c

d

-

d

c

b

a

A

B

C

D

which gives the relations

D = 10 + d - a (as a > d)

C = 10 + c - 1 - b = 9 + c - b (as b > c - 1)

B = b - 1 - c (as b > c)

A = a - d

for those numbers where a>b>c>d.

A number will be repeated under Kaprekar's operation if the resulting number ABCD can be written using the initial four digits a,b,c and d. So we can find the kernels of Kaprekar's operation by considering all the possible combinations of {a, b, c, d} and checking if they satisfy the relations above. Each of the 4! = 24 combinations gives a system of four
simultaneous equations with four unknowns, so we should be able to solve this system for a, b, c and d.

It turns out that only one of these combinations has integer solutions that satisfy 9 ≥ a ≥ b ≥ c ≥ d ≥ 0. That combination is ABCD = bdac, and the solution to the simultaneous equations is a=7, b=6, c=4 and d=1. That is ABCD = 6174. There are no valid solutions to the simultaneous equations resulting from some of the digits in {a,b,c,d}
being equal. Therefore the number 6174 is the only number unchanged by Kaprekar's operation — our mysterious number is unique.

For three digit numbers the same phenomenon occurs. For example applying Kaprekar's operation to the three digit number 753 gives the following:

753 - 357 = 396
963 - 369 = 594
954 - 459 = 495
954 - 459 = 495

The number 495 is the unique kernel for the operation on three digit numbers, and all three digit numbers reach 495 using the operation. Why don't you check it yourself?

How fast to 6174?

It was about 1975 when I first heard about the number 6174 from a friend, and I was very impressed at the time. I thought that it would be easy to prove why this phenomenon occurred but I could not actually find the reason why. I used a computer to check whether all four digit numbers reached the kernel 6174 in a limited number of steps. The program, which was about 50 statements in Visual
Basic, checked all of 8991 four digit numbers from 1000 to 9999 where the digits were not all the same.

The table below shows the results: every four digit number where the digits aren't all equal reaches 6174 under Kaprekar's process, and in at most seven steps. If you do not reach 6174 after using Kaprekar's operation seven times, then you have made a mistake in your calculations and should try it again!

Iteration

Frequency

0

1

1

356

2

519

3

2124

4

1124

5

1379

6

1508

7

1980

Which way to 6174?

My computer program checked all 8991 numbers, but in his article Malcolm Lines explains that it is enough to check only 30 of all the possible four digit numbers when investigating Kaprekar's operation.

As before let's suppose that the four digit number is abcd, where

9 ≥ a ≥ b ≥ c ≥ d ≥ 0.

Let us calculate the first subtraction in the process. The maximum number is 1000a+100b+10c+d and the minimum number is 1000d+100c+10b+a. So the subtraction is:

The possible value of (a-d) is from 1 to 9, and (b-c) is from 0 to 9. By running through all the possibilities, we can see all the possible results from the first subtraction in the process. These are shown in Table 1.

Table 1: Numbers after the first subtraction in Kaprekar's process

We are only interested in numbers where the digits are not all equal and

a ≥ b ≥ c ≥ d,

therefore we only need to consider those where (a-d) ≥ (b-c). So we can ignore the grey region in Table 1 which contains those numbers where

(a-d) < (b-c).

Now we arrange the digits of the numbers in the table in descending order, to get the maximum number ready for the second subtraction:

Table 2: Maximum numbers, ready for the second subtraction

We can ignore the duplicates in Table 2 (the grey regions), and are left with just 30 numbers to follow through the rest of the process. The following figure shows the routes which these numbers take to reach 6174.

How these 30 numbers reach 6174

From this figure you can see how all the four digit numbers reach 6174 and reach it in at most seven steps. Even so I still think it is very mysterious. I guess Kaprekar, who discovered this number, was extremely clever or had a lot of time to think about it!

Two digits, five digits, six and beyond...

We have seen that four and three digit numbers reach a unique kernel, but how about other numbers? It turns out that the answers for those is not quite as impressive. Let try it out for a two digit number, say 28:

It doesn't take long to check that all two digit numbers will reach the loop 9→81→63→27→45→9. Unlike for three and four digit numbers, there is no unique kernel for two digit numbers.

But what about five digits? Is there a kernel for five digit numbers like 6174 and 495? To answer this we would need to use a similar process as before: check the 120 combinations of {a,b,c,d,e} for ABCDE such that

9 ≥ a ≥ b ≥ c ≥ d ≥ e ≥ 0

and

abcde - edcba = ABCDE.

Thankfully the calculations have already been done by a computer, and it is known that there is no kernel for Kaprekar's operation on five digit numbers. But all five digit numbers do reach one of the following three loops:

As Malcolm Lines points out in his article, it will take a lot of time to check what happens for six or more digits, and this work becomes extremely dull! To save you from this fate, the following table shows the kernels for two digit to ten digit numbers (for more see Mathews Archive of
Recreational Mathematics). It appears that Kaprekar's operation takes every number to a unique kernel only for three and four digit numbers.

Digits

Kernel

2

None

3

495

4

6174

5

None

6

549945, 631764

7

None

8

63317664, 97508421

9

554999445, 864197532

10

6333176664, 9753086421, 9975084201

Beautiful, but is it special?

We have seen that all three digit numbers reach 495, and all four digit numbers reach 6174 under Kaprekar's operation. But I have not explained why all such numbers reach a unique kernel. Is this phenomenon incidental, or is there some deeper mathematical reason why this happens? Beautiful and mysterious as the result is, it might just be incidental.

Let's stop and consider a beautiful puzzle by Yukio Yamamoto in Japan.

If you multiply two five digit numbers you can get the answer 123456789. Can you guess the two five digit numbers?

This is a very beautiful puzzle and you might think that a big mathematical theory should be hidden behind it. But in fact it's beauty is only incidental, there are other very similar, but not so beautiful, examples. Such as:

(We can give you a hint to help you solve these puzzles, and here are the answers.)

If I showed you Yamamoto's puzzle you would be inspired to solve it because it is so beautiful, but if I showed you the second puzzle you might not be interested at all. I think Kaprekar's problem is like Yamamoto's number guessing puzzle. We are drawn to both because they are so beautiful. And because they are so beautiful we feel there must be something more to them when in fact their beauty
may just be incidental. Such misunderstandings have led to developments in mathematics and science in the past.

Is it enough to know all four digit numbers reach 6174 by Kaprekar's operation, but not know the reason why? So far, nobody has been able to say that all numbers reaching a unique kernel for three and four digit numbers is an incidental phenomenon. This property seems so surprising it leads us to expect that a big theorem in number theory hides behind it. If we can answer this question we
could find this is just a beautiful misunderstanding, but we hope not.

Note from the editors: many readers noticed that repeatedly adding up the digits of any of the kernels of Kaprekar's operation always equals 9. Find out why in this follow-up to the article.

References

Lines, Malcolm E., A number for your thoughts: facts and speculations about numbers..., Bristol: Hilger (1986)

Nishiyama, Yutaka, Kurashi no Algorithm, Kyoto: Nakanishiya (1993)

About the author

Yutaka Nishiyama is a professor at Osaka University of Economics, Japan. After studying mathematics at the University of Kyoto he went on to work for IBM Japan for 14 years. He is interested in the mathematics that occurs in daily life, and has written seven books about the subject. The most recent one, called "The mystery of five in nature", investigates, amongst other things, why many
flowers have five petals. Professor Nishiyama is currently visiting the University of Cambridge.

Comments

When I was at college in the '80 in Derbyshire this was something our computer studies teacher talked about, he had read an article in a forth magazine about this and described it as being 6174 as the center of a series of concentric rings when plotted as a scatter diagram.

After listening to him prattle on about it for so long I eventually set out to prove that he didn't know what he was talking about (not unusual). So on the first day I wrote a program to calculate the route to 6174 for all 4 digit numbers, unfortunately I only had 32k of ram on my home machine and no floppy and I also needed 20k for high resolution graphics. Clearly since any calculation with the same digits would yield the same result I used 4 different FOR loops to reduce the storage requirement.
so
for a= 9 to 0
for b= a to 0
for c= b to 0
for d =c to 0
This allowed me to store and plot the results and it was nothing like as described. Apparently, my teacher explained the scatter diagrams had been rotated to make the plot of concentric rings, LOL

Anyway I said to my teacher that next I would plot the steps for all 4 digit numbers as the start of my proof investigation, when I came into college the following day my teacher said that he has passed my original ideas to the high level course and my work was a waste of time. I pointed out that in their plot they found that most numbers were not a route to 6174 and so were a jumble this as I had discussed the problem with my father who pointed out that the result of this operation is always a multiple of the base -1. Again I was finding storage a problem so I used the model discussed in the forth magazine to keep things simple, i.e. order digits high to low then reading left to right subtract 4rd digit from 1st and 3rd from second so for 7641 that would give the model 62. All numbers with model 62 in base 10 give 6174 as a result in 1 operation. Other models also share the same number of steps to reach that number. So plotting just the models you can see a much smaller graph and this moved me closer to being able to say how many steps would be required before reaching 6174.

So my question is, has anyone else produced the proof for this and answered why the numbers act this way to reach 6174 and also why 5 digits for example make a ring instead of a single number? after my experiences with my teacher I haven't published or discussed my own work on this with anyone else.

I read about this in Parc, S., “50 Visions of Mathematics”, Oxford University Press, (2014) and I wondered if this was true for any number base rather than just base 10.
On trying three digit numbers in octal and hexadecimal I found that they each had an equivalent kernel. This led me to solve the equations for finding the kernel, but using a variable, β, for the number base rather than restricting it to decimal numbers.

The results were as follows:
For two digit numbers there is a kernel if the base is of the form 3*(n – 1) + 5
For three digit numbers there is a kernel if the base is even i.e. of the form 2*n
For four digit numbers there is a kernel if the base is of the form 5*n, though there are kernels for base 2 and base 4.
For five digit numbers there are kernels for bases of the form 3*n

As I was using the computer algebra add-on for Microsoft Word I chose not to solve individually the hundreds of equations for six digit numbers. However, solving the equations based on the base-ten six digit numbers I did find that there are kernels for bases of the form 9*n + 8, 2*n, 15*n + 10, and 2*(n + 2).
In the above n takes the values 0, 1, 2, ...

I haven’t a proof, but it does seem likely that for any value with 3*n digits and an even base, there is a kernel which has n digits of value β – 1, β/2, and β/2 – 1. This is true for three digit numbers, six digit numbers, and some nine and twelve digit numerical examples I’ve tried.

I’ve not seen any references to this so I thought I would post these results as they may be of interest to other readers.

I made some brute force computing for test of four digit numbers with different bases up to 40. The principle for presentation is extrapolation of the method used to present HEX code: A = 10, B = 11 ….. F = 15, G = 16 … Z = 35, [ = 36, … ^ = 39, _ = 40.

I found that only bases 5, 10 and 40 terminate to one value: 3032 (=dec 392), 6174 and O7VG (= dec 1548456) respectively with max number of loops 4, 7 and 21 respectively.

When it comes to base = 20, loop terminates to C3F8 (= dec 97508) for many start values with max number of loops = 10
Base 15, 30, 35 do not terminate to one value.

Below a sample of loops for different bases. “…” means that loop is unterminated and will continue for ever. “…” is printed after a value in the last position N when it has been detected earlier in the loop in position N-2 or earlier. E.g. the loop
0045 5311 4132 3043 3552 3133 1554 4042 4132 ...
will continue for ever like
0045 5311 4132 3043 3552 3133 1554 4042 4132 3043 3552 3133 1554 4042 4132 3043 3552 3133 1554 4042 4132 etc

All bases are not included in this comment, but are produced in the program. I’m thinking of extending the number of digits to be variable too…
All bases have been given a random start value within the definition area, and 10 loops are printed. All loops are not presented in this comment.

Let's omit the first stage of Kaprekar's algorithm and just look at what happens when the only rearrangement is reversing the digits in three of the numbers he looks at and doing a repeated operation.

Find the absolute difference between a number and its reverse, and then the difference between that and its reverse, and so on.

2005, 2997, 4995, 999

1789, 8082, 5274, 549, 396, 297, 495, 99

6174, 1458, 7083, 3276, 3447, 3996, 2997, 4995, 999

The absolute difference between the reverses of these repnines gives us our kernel, which is zero. That's true of all palindromes of course, though the operation in the algorithm producing a palindrome won't necessarily lead to a kernel. For example, consider summing them instead

Add the kernel 495 to its reverse, 594, and you get another famously mysterious number 1089. Do the same with 6174 and get 10890, (which is also the result of R&A 1089). R&A 549945 = 1099890, R&A 631764 = 1098900.

Does it surprise anyone that if we treat all numbers from 0 (0000) as four digits up to 9999 we end up at 6174 (excluding the series where numbers are the same), which is pretty darn close to the golden ratio when divided by the whole sample number (0.6180).

Hi,
The maximum common divisor of 495 and 6174 is 9.
In the 9 scale you can find all the 1467,1476,4167,4617 .. etc iterations.
In a magic scope: +495 + 6174 = 6669. +6669/9 (MCD)= 714. 714 is considered lucifer number.
Best,
Martin

Some thoughts
*11+22+33+44+55+66+77+88+99=495
*111+222+333+444+555+666+777+888+999+1110+69=6174
*6174+495=6669
*Square root of (1467,1476,1647,1674....) Only hole numbers 42 and 69. 42*42=1764 69*69=4761
* 69+42=111
* 0000,1111,2222,3333,4444,5555,6666,7777,8888,9999. 6174 should have a property similar from the rest of them.
(1467,1476,1674,1647 ...etc)/5555 -average of above, 6174 is 1,111. If variations are in order, 1,111 is in the middle of the variations.
* All numbers in Kaprekar process are divisible by 9.

Here below is sequence 7641, an exact 4 digit sequence in the famous repeating 24 digit Digital Root Fibonacci Sequence. Below, I'll use a date utilizing a special date formula to arrive at a 2 digit result, but a 4 digit result is more interesting; this number was not expected; it's just the result of some inspired calculation methods.

Do the same for 7-27-2017, and a 4 digit sequence result is 5628, also exact.

With both 4 digit numbers, those are 8 in exact order in the 24 repeating digit sequences in DR FS.

7641 and 5628 can be translated to English letters (A=1, B=C....J_10 as DR 1, Z_26 = as DR 8 etc).
Since F=6, O=15, and X=24, 6 above can be F, or O or it an be X...
7 can be G, P or Y...(or PIG instead of PYG) since letter I is dr 9 and does not taint the calculation...

7641 5628
GODA EOBH

Likewise, adding an R which is letter 18 which is DR 9, also does not taint the calculation.

And, when you re-arrange those letters, you can see this result:

"HE GOD BOAR"

Just number fun?.... Is this all just for our amusement or for or edification?

Look at 6 and 9 digit numbers.. 6 and 9 are 2nd and 3rd multiples of 3..
Moreover, the results for 6 an 9 digits are all containing the numbers 495 or 6174 and then taking the 6 digit versions which add extra digits by putting the original numbers backwards interspaced with itself.. For 495, _4 _9_5 woven with 5_9_4_ becomes the mirrored number 549945…. and for 6174 then 6_17_4 and the added two digits 3 and 6 only _3__6_ which is only one off from backwards first and last digits of 4 and 6 becoming 631764.
Now for the related NINE digit versions, you take the six digit versions and stutter the numbers as such
549945 becomes 5_49_94_5 adding the backwards of 495 interspersed as _5__9__4_ to become 554999445 and
864197532 contains _641_753_. or 641753 which is 631764 only shifted up or down one for three of the digits.

Then the EIGHT and TEN digit numbers are related somehow to the 4 and 6 digits numbers for at least the first answer which is the same as the eight digit number answer just stuttering in another mirrored copy of the digits 3 and 6 again. The middle number unsure how that might derive till. The last ten digit answer clearly contains the last eight digit answer as such: 9975084201. =. _9750842_1 ….

And don't get me started on the patterns for the added digits and evens and odds… ;)

Your link to "Mathews Archive of Recreatrional Mathematics" is no longer good. The closest related page I could find by Googling was the one here:

In the table showing the different digits a pattern involving 495 and 6174 emerges in the higher numbers. For 6, 8, 10 digits they can all end at 6174 with a different number of 3's and 6's in the middle. 6 and 9 digits can end with 495 rearranged 2 then 3 times respectively